In a YDSE a parallel beam of light of wavelength 6000Å is incident on slits at angle of incidence `30^(@)`. A and B are two thin transparent films each of refractive index 1.5. thickness of A is `20.4mum`. Light coming through A and B have intensities I and 4I resepctively on the screen. Intensity at point O which is symmetric relative to the slits is 3I. The central maxima is above O.
(i). what is the maximum thickness of B to do so.
Assuming thickness of B to be that found in part (i)
Answer the following parts.
(ii). Find fringe width, maximum intensity and minimum intensity on screen.
(iii). Distance of nearest minima from O
(iv) intensity at 5 cm on either side of O.

Answer

(a) `t_(B)=120 mum`
(b) `beta = 6 mm ; I_(max)=9I,I_(max)=I`
(c) `beta//6 =1 mm` (d) `I (at `5 cm` above `O`)`=9 I`
`I("at" 5 cm "below" O)=3I`

Explanation

(a) For maxima thickness of `B.` the central maxima is just above `O`.
`rArr` Phase difference at `O` must be `=(2pi)/(3) {therefore cos phi=-(1)/(2) "for" 3I "intensity"}`.
`rArr` Path difference at `O=(2pi)/(3)xx(lambda)/(2pi)=(lambda)/(3)`
`rArr d sin 30^(@) +(mu_(A)-1)xxt_(A)-(mu_(B)-1)t_(B)=lambda//3`
get `t_(B) =120 mum`.
(b) Now `t_(B)=120 mum`.
`bete=(lambdaD)/(d)=(6000xx10^(-10)xx1)/(0.1xx10^(-3))=6 mm` .
`I_(max)=(sqrt(4I)+sqrt(I) )^(2)=9 I`
`I_(min)=(sqrt(4I)-sqrt(I))^(2)= I`
(c) `DeltaP` path difference at `'O'`
`rArr DeltaP=dsintheta+(mu_(A)-1)t_(A)-(mu_(B)-1)t_(B)`
`=0.1 sin30^(@)+(1.5-1)xx20.4-(1.5-1)xx120`
`DeltaP=2xx10^(-7)m`
`therefore y(d)/(D)=DeltaP`
`thereforey=(D)/(d)DeltaP=(1)/(0.1xx10^(-3))xx2xx10^(-7)=y=2 mm`
`thereforebeta=(lambdaD)/(d)=6mm`
distance between maxima and minima `=(beta)/(2)=3mm`
distance of nearest minima from `'O'=(beta)/(6)=1mm`

(d) So phase difference at `P_(1)`
`Deltaphi=(yd)/(D)xx(2pi)/(lambda)`
`=(48xx10^(-3)xx10^(-4))/(1)xx(2pi)/(6000xx10^(-10))`
`Deltaphi=16pi`
Intensity at `'P'_(1)`
`I_(1)=4I+I+2sqrt(4I^(2)) cos (16pi)`
`I_(1)=9 I`
Intensity at point `'P'_(2)`
phase differnce at `'P'_(2)`
`DeltaP=(2pi)/(lambda)xxDeltaP`
`DeltaP=(yd)/(D)=(52mmxx0.1mm)/(1m)`
`Deltaphi=(2pi)/(6000xx10^(-10))xx(52mmxx0.1mm)/(1m)`
`Deltaphi=52(pi)/(3)=16pi+4pi//3`
`I_(2)=4I+I+2sqrt(4I^(2)) cos (16pi+4pi//3)`
`I_(2)=4I+I+2sqrt(4I^(2))cos (4pi//3)`
`=4I+I-2xx2Ixx(1)/(2)`
`I_(2)=3I`